A GENESIS FOR COMPACT CONVEX SETS

APPROVED:

Major Professor f

y&rvu# /£• ^Lnor Professor //

JT I rector of the Department of Mathematics

Dean of the Graduate School

A GENESIS FOR COMPACT CONVEX SETS

THESIS

Presented to the Graduate Council of the

North Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

1 MASTER OF SCIENCE

By

Ronald D. Ferguson, B. A.

Denton, Texas

May, 1969

TABLE OF CONTENTS

?

INTRODUCTION iv

Chapter

I. CONVEX HULLS t. . 1

II. SUPPORT THEOREMS 16

III. EXTREMAL CHARACTERIZATION OF CONVEX SETS 26

IV. A PROPOSITIONAL SUBSET OF THE GENESIS

OF A SET 35

APPENDIX 42

BIBLIOGRAPHY. 43

xix

INTRODUCTION

This paper was written in response to the following

question: what conditions are sufficient to guarantee

that if a compact subset A of a topological linear space . t

L3 is not convex, then for every point x belonging to

the complement of A relative to the convex hull of A

there exists a line segment yz such that x belongs to

yz and y belongs to A and z belongs to A? Restated in •

the terminology of this paper the question may be given

as follows: what conditions may be imposed upon a

compact subset A of L3 to insure that A is braced?

The pursuit of this problem gave rise to the

concept of the genesis of a compact convex set. The

class of braced sets having convex hull A could be

considered as a propositional subset of the genesis

of A.

In general, the terminology and language of this

paper is that followed by Valentine (1) in his book on

convex sets. Terms that are newly introduced herein

are marked by an asterisk.

i v

CHAPTER I

CONVEX HULLS

Convex sets were first studied by Brunn in 1887.

Later, Minkowski added considerably to the body of

knowledge that had begun to accumulate around convex

sets. As a preliminary to the consideration of some

theorems related to convex sets the following definitions

are introduced:

Definition 1.1 Suppose S is a set of elements

with a collection of subsets called open sets which

satisfy the following axioms:

(a) The union of each collection of open sets

is an open set.

(b) The intersection of each finite collection

of open sets is an open set.

Then S together with its collection of open sets is

called a topological space. The collection of open sets

is called the topology of S. It can readily be deduced-

from (a) and (b) that <j> and S are open sets. Henceforth,

the better known properties of topological spaces will

be assumed.

Definition' 1.2 The sets C£S and D£S are

complementary iff CUD=S and CftD=<j>.

The complement of A relative to B is the set

denoted by B^A={x| xeB and x^A }. Obviously if C and

D are complementary, then C=S^D. C denotes the

complement of C, i.e. C=S^C=D.

Definition 1.3 A closed set in a topological

space S is the complement of an open set, i.e. A is

closed iff A=SM3, where B is open.

Definition 1.4 A linear space I over the field

of real numbers R is a space for which there exist two

binary operations, vector addition ( indicated by + )

from £*£ into I and scalar multiplication ( indicated

by juxtaposition ) from Rx£ into I which satisfy the

following axioms:

If x,y,z e I and a,3 e I then

(1) x+y e I (4) ax e I

(2) x+y=y+x (5) a(x+y)=ax+ay

(3) (x+y)+z=x+(y+z) (6) a($x) = (a3)x

(7) (a+$)x=ax+3x

(8) for every pair x,y e I, there exists a z e I

such that x+z=y

(9) if 1 is the identity of the real field R, then

lx=x for every x e I.

Number (8) implies the existence of an additive

identity for vector addition since x=y is a possible

pair. 0 will be used to denote the additive identity

and is synonomous with the origin of a vector space.

Definition 1.5 Let ll be a linear space with a

topology. Let x e I. Then an open set N in the topology

of £ such that x e N is called a neighborhood of x and

is denoted by N(x).

Definition 1.6 Let S be a set in I. If x e S

then x is an interior point of S iff there exists a

neighborhood N(x) of x such that N(x)^S. The set of

all interior points of S is called the interior of S

and is denoted by intS.

It is well known that intS=S iff S is an open set.

Definition' 1.7 Let S be a set in £ and x e t.

Then x is said to be a limit point of S iff for every

neighborhood N(x) of x, N (x) SS^x}^.

Definition 1.8 Let S be a set in £. Then S'

denotes the set of all limit points of S and is called

the derived set of S. SuS1 is called the closure of

S and is denoted by clS.

One may easily verify that clS is a closed set.

Definition 1.9 Let S be a set in I. The set

of points x e £ such that for every neighborhood N'(x)

of x, N(x)ftS (j) and N(x)ftS <f> is called the boundary of

S and is denoted by bdS.

4

A boundary point of S is said to be isolated if

it is not a limit point of S.

Definition 1.10 Let f be a function which maps

a subset S in a tbpological space % into a set S* of a

topological space £*. The function f is said to be

continuous at a point x e S iff for each neighborhood

V[f(x)] of f(x) in 5* there exists a neighborhood U(x)

of x in £ such that if y e SflU(x), then f(y) e V; i.e.,

f-(snu) £V.

Definition 1.11 If a linear space £ has a topology

then the topology is Hausdorff iff for each x e £,

y e H, x^y, there exist two disjoint open sets containing

x and y respectively.

Definition 1.12 A topological linear space L is

a linear space with a Hausdorff topology such that the

operations of vector addition, f(x,y)=x+y denoted by

(x+y)/(x,y), and scalar multiplication, f(a,x)=ax denoted

by (ax)/(a,x), where x e L , y e L , a e R are continuous

in all variables jointly.

Definition 1.13 If x e £, y e I, the line segment

xy joining x and-y is the set of all points of the form

otx+(l-a)y where a e R , 0<a<l.

Definition 1.14 A set is said to be star-shaped

relative to a point x e I iff for each point y e S

xy?S.

Definition 1.15 A topological linear space L

is locally star-shaped iff for each neighborhood U(0)

of the origin 0 of L, there exists a neighborhood V(0)

of 6 such that V(9)£U(0) and such that V(0) is star-shaped

relative to 0.

Definition 1.16 If A is a set in £ and if x e I,

then the set A+x={y+x| y e A } is called the translate

of A by x.

Definition 1.17 If A is a set in I and a e R

aA={ax| x e A } is called the scalar multiple of A by a.

Several well known theorems are tacitly assumed

in this paper in order to devote time to properties

and theorems whose consequences have a greater bearing

on the subject under consideration. So it is without

apology that the following theorem is offered without'

proof:

Theorem 1.1 The continuity of vector addition

and scalar multiplication implies that each translate

and non-zero scalar multiple of an ope« set is open. tJ

Theorem 1.2 A topological linear space L is

locally star-shaped and locally symmetric.

Proof: Let U be a neighborhood of the origin 0.

Now ax/(a,x) is continuous at (0,0), so that by the

definition of continuity there exists a neighborhood

V(0) of 0 and a number 6 >0 such that if |a|<6 and if

x e V(0), then ax e U(0). Immediately, aV?U whenever

|a|<6. Consider the class T={aV| |a|<6 }. Then

by definition of a topology and neighborhood uT is

a neighborhood of 8. Also, N= uT£U(0), and N is

star-shaped and'symmetric relative to 0.

Theorem 1.3 If Ln is an n-dimensional topological

linear space, then LD is linearly isomorphic with

Euclidean n-space En. This means there exists a n n one-to-one linear bicontmuous map of L onto E .

The immediate import of this theorem is that theorems

n n holding in L are immediately applicable to E and

vice-versa. For a proof of Theorem 1.3 see Valentine

(1, pp. 7-8) .

Definition 1.18 A set S££ is convex iff for each

pair of points x,y e S, xy^S.

Definition 1.19 A topological linear space L is

locally convex iff for each neighborhood U(6) of 6 there

exists a convex neighborhood V(0) of 0 such that V(8)£U(0),

An obvious alternative to Definition 1.18 is that

a set S i s convex iff it is star-shaped relative to

every point x e S. Similarlyf Definition 1.19 could be

altered to conform with Definition 1.15.

Definition 1.20 A convex set in L which has an

interior point is called a convex body.

Theorem 1.4 The intersection of any collection of

convex sets in a linear space £ is a convex set.

Proof: Let C be a collection of convex sets

in I. Consider fiC. Suppose QC has two points x and y.

Then x and y belong to every element of C. But this

implies that xy?;A for every A e C, so that xy£f2C. Then

ftC is convex. If QC does not have at least two points,

then the conclusion is obvious.

Definition 1.21 The convex hull of a set SsL is

the intersection of all convex sets in L containing S

and is denoted by convS.

S is convex iff convS=S.

Definition 1.22 A set S of a topological linear

space L is bounded iff for each neighborhood N(0) of 0

there exists a positive scalar multiple of N(0), aN(0),

such that S£aN(0).

Definition 1.23 If A is a subset of S where S

is a topological space, then A is compact iff every

open covering of A has a finite subcover. 'In other words,

if T is the topology of S, A is compact iff for every

class of open sets C£T such that A^uC, there exists a

finite subset B of C such that A£uB.

It is well known that A£Lnis compact iff A is

closed and bounded.

• Definition 1.24 A flat or linear variety of a

linear space SL is a translate of a linear subspace of

8

I. Two flats are parallel iff one is a translate of

the other; i.e., if F and F* are flats in I, then F and

F* are parallel iff there exists an x e I such that

F=F*+x. Further, the dimension of.a flat is the

cardinality of the basis of the corresponding parallel

linear subspace. A flat of dimension 1 is called a

line.

Any good book on linear algebra can provide

information concerning the basis and dimension of a

vector space.

Theorem 1.5 Each finite dimensional flat of a

topological linear space L is closed.

Proof: Suppose that V is a finite dimensional

flat of L. Assume, by way of contradiction, that

x e clVM#<f>. Then there exists a flat of minimal

dimension containing V and x. Let W be such a flat.

Describe a relative topology on W as follows: let

T={WftU | U e T*, where T* is the topology of L }.

Now if x is in the closure of V relatiye to W, Theorem 1.3 • i

implies that x e V, a contradiction. Thus, there exists

a neighborhood N(x) of x relative to W such that N(x)ftV=(j>.

But by definition of T there exists a neighborhood of x,

U (x) e T* such that U(x)ftW£N(x). Since V£W, U(x)ftV=<j>.

But then x £ clV. Thus clV'vV=({) and V is closed.

The next theorem relates a convex set to its line

segment subsets. For a proof see Valentine (1, p.16).

Theorem 1.6 Let S be a set . in the n-dimensional

linear space &n. Let be defined recursively as

follows:

Sx= u xy S.= u xy

y«i x ^ i " 1

y si-i in~l i n

Then = convS if in satisfies the inequality 2 <n+l<2 n. n Theorem 1.7 If A and B are two convex sets in £,

then conv(AuB)= u [aA+(l-a)B] = D. 0£a£l

Proof: If x e A, y e B, then ax+(l-a)y e conv(AuB)

when O^oKl. Thus D^conv(AuB) immediately. Now suppose

z e D, w e D. By definition of D there exist points

x,x* e A and y,y* e B such that z e xy, w e x*y*. But

xx*£A, yy*CB, so that conv(xx*uyy*)CD. Then zw is a

subset of conv(xx*uyy*) so that zw?;D. Thus D is convex

and (AoB)£D. Immediately, D=conv(A B).

Definition 1.25 Let SQl. The kernel K of S is

the set of all points z e i such that zx^S for every

x e S. In other words, the kernel K of S is the set of

all those points with respect to which S is star-shaped.

Notice that S is convex iff S=K where K is the kernel of

S. Also, note that K is itself a convex set follows

immediately from the definition of K, so that K is often

referred to as the convex kernel of S.

10

Definition 1.26 The interior of a set ScL

relative to the minimal flat containing it is denoted

by intvS. For a linear space SL, if x e I, y e I,

x^y, then intvxy denotes the set of all points of the

form ax+(l-a)y, where 0<a<l. If x e I, then intvx=x.

Definition 1.27 A point of x e S is a core point

of S iff for each point y el, with y^x, there exists a

point z e. intvxy such that xz£S. The set of all such

core points of S is called the cores.

Theorem 1.8 If S£L is star-shaped relative to a

point p, then the clS is star-shaped relative to p.

Proof: Let y e clS. Choose some point of yp,

say Ay+(1-A)p where 0<^£1. Now, Ax+(1-A) p/(x) is continous

in x, so that for each neighborhood U of Ay+(1-A)p in L

there exists a neighborhood V(y) of y such that

Ax+(1-A)p e U whenever x e V(y). Since y e clS, let

x e VfJS. Then Ax+(1-A)p e UftS. But this implies that

Ay+(1-A)p e clS. Then yp^clS.

Corollary 1.8 Let C be a convex set in L. Then

the closure of C is convex.

Proof: Let z e clC, w e clC. As in Theorem 1.8

Ax+(1-A)y/(x,y) is continuous, so that if some point of

zw is chosen, say Az+(1-A)w where 0<A<1, then, as in

Theorem 1.8, zw£clC, so that clC is convex.

11

1.8 implies immediately that

as a result cconvS£cl(convS).

Dsed convex set which contains S.

as cl (cojivS) ccconv. Therefore,

Theorem 1.9 If S is a set in a topological

linear space L, then the closed convex hull of S, denoted

by cconvS is the same as the closure of the convex hull

of S.

Proof: Corollary

cl(convS) is convex, and

Note that cconvS is a clc

Hence, convS^cconvS. Thi

cl(convS)=cconvS.

Theorem 1.10 Supbose S is a set in L and S is

star-shaped relative to ja, point p. If p e intK, where

K is the convex kernel 0]E S, and if y e clS, then

intvyp^intS.

Proof: Let.y e clS, and without loss of generality

let y=4>. Now all that is necessary is to show that

apcints for all 0<os<l since y=<f>. Since p e intK, there

exists a neighborhood V(<f>) of <f> such that p+V(<J>) £intK.

Then, for 3=-a/(l-a) , by Theorem 1.1, the set 3v() is

open. Since e clS, there exists a point u e $[V(c)>) ] QS.

Thus, there exists a point v e V(tj>) such that $v e S.

is an open set contained in S,

Also, x=(1-a)6v+a(p+V)=ap and

The set U=(1-a)Bv+a(p+V)

since v e S, p+V^intK.

ap e U?intS. Thus intvypsintS.

12

Corollary 1.10 Let C be a convex set in a linear

topological space L. If x e intC, y.e clC, then

intvxycintc.

Proof: Since intC=intK, Theorem 1.10 implies that

intvxy^intC immediately.

Theorem 1.11 Let S be a set in L and suppose

that S is star-shaped relative to a point p. Then, if

p e intS; the intS is star-shaped relative to p.

Proof: Let y e intS and without loss of generality

suppose that p=9, and choose x=ay, 0<a<l. Since

y e intS, there exists a neighborhood U(y) of y such that

U(y)£intS. Since p e K, aU(y)£S, and since aU(y) is open,

ay e aU(y)£intS, for 0<a<l. Thus yp£intS, and intS is

star-shaped relative to p.

Corollary 1.11 If S is a convex set in L, then

intC is convex.

Proof: The proof is immediate from Theorem 1.11

since C is star-shaped relative to every point in intC.

Theorem 1.12 If S^L, then intS^coreS.

• Proof: Suppose z e intS. Let U be a neighborhood

of z with U^intS. Let y e L^{z}. Since F(a)=z+a(y-z) is

continuous at a=0, there exists a 6 such that if 0<a<6, r S=3 *

then x=F(a) e U. Thus x e intvzy and xy£U£intS. Then

z e cores.

13

Theorem 1.13 If C is a convex set in I, then

coreC is convex.

Proof: Let x e coreC, y e coreC. Hence y e C

and C=K, where K is the convex kernel of C. Without

loss of generality assume that y=0. Then, let z=ax,

where (Ka<l. Consider any vector u where u^0. Since

x e coreC, there exists a constant <5>0 such that

x+gu £ C for 0<s3<6. C is star-shaped relative to y=0

so that a(x+3u)^C for 0<;3<6. But this implies that

ax e coreC and thus xy^coreC. Then coreC is convex.

Theorem 1.14 If C is a convex body in L, then

coreC=intC=int(clC).

Proof: Let y e coreC and x E intC. Since

y e coreC, there exists a point z such that y e intvxz

and such that zy£C. Hence, z-e clC. By Corollary 1.10

intvxz^intC, so that y e intC. Consequently, coreC^intC^

Theorem 1.12 implies that intC^coreC; hence, intC=coreC.

Now, let y £ int clC, x e intC. There exists a point

z £ clC such that y £ intvzx. Again by Corollary 1.10,

intvzx<;intC. Hence, y £ intC so that int clC^intC.

Since C^int clC, then int clC=intC.

Theorem 1.15 If S is an open set in L, then

convS is open.

Proof: Corollary 1.11 implies that int convS is

convex. Since S^convS, and Sft(bd convS) =<p, then

14

Spirit convS. But int cortvS is convex, so that the

previous sentence implies convS£int convS. Also,

int convS£convS, so that convS=int convS. Thus,

convS is open.

It should be pointed out that the convex hull

of a closed set need not be closed. As an example con-

sider the the set S in Euclidean two-space E2 with

rectangular coordinates (x,y) defined by

S={(x,y)| x2y2=l, 0<y<«> }

S is closed but convS={(x,y)| y<0 } is not closed.

CHAPTER BIBLIOGRAPHY

1. Valentine, Frederick A., Convex Sets, New York, McGraw-Hill Book Company, 1964.

15

CHAPTER II

SUPPORT THEOREMS

The concepts of 'hyperplanes and half-spaces

are very useful in the formulation and proof of various

theorems concerning convex sets. As soon shall become

apparent, hyperplanes are related to flats in a very

intimate manner. In order to define a hyperplane the

notion of a linear functional is introduced.

Definition 2.1 Recall that a real valued function

is often called a functional. A linear functional is

a function from & to R which is additive and homogeneous,

that is

f(x+y)=f(x)+f(y) x £ y e &

f(ax)=af(x) x £ a e r.

Definition 2.2 Suppose HC&, then H is said to be

a hyperplane iff there exists a non-identically zero

linear functional f and a real constant ot e r such that

H={ x e £| f(x)=a } denoted by H= [f ;<*] .

Theorem 2.1 A hyperplane is a flat.

Proof: Suppose that f is a linear functional

and that f(x)£o, x e I. Let H=[f:0t]. Since ffo, let

16

17

z £ £, with f(z)^0. Choose an arbitrary point q e It

and since f(z)j^0, define p= [-f (q)/f (z) ] (z)+y where

y e H. Since f is linear,

f (p)=f{ [ — f?(q) /f (z) 3 (z)+y}

f(p) =-f(q)+f(y) so that

f(p)+f(q)=f(p+q)—f(y)=a.

Thus p+q e H. But this statement together with the

definition of p implies that &=H+Rz. Consequently,

H is a flat.

Definition 2.3 If f is a linear functional and

if AC&, then f(A)>a means f(x)>a for every x e A.

Similar definitions may be made for inequalities that

are strict or reversed.

Definition 2.4 The hyperplane H=[f:a] bounds

the set A££ if either f(A)>a or f(A)<a holds.

Definition 2.5 The sets {x e H| f(x)>a },

{ x e £| f(x)<a }, { x e £ | f(x)!>a }, and { x £ 11 f(x)<ot }

are called half-spaces of I.

Axiom 2.1 A partial order on a set S is a subset

0 of SxS such that if (x,y) e 0, then (y,x) £ 0, and if

(x,y) e 0 and (y,z) e 0 then (x,z) e 0. Suppose that

S is a partially ordered set, then S contains at least

one maximal linearly ordered subset, where a linear order

is a partial order with the additional condition that

(x,y) i 0 implies that (y,x) e 0.

18

Examples of partially ordered relations are

less than or equal and set inclusion "5." Zorn's

maximal principle is useful in the proof of a number of

theorems concerning convex sets. It is equivalent to

the axiom of choice and also to the Hausdorff maximal

principle. For a proof of this equivalence and a

discussion of related topics see Kelley (1, pp. 31-36).

Zorn's Maximal Principle If T is a partially

ordered set and each linearly ordered subset of T has

an upper bound in T, then T contains at least one

maximal element.

Theorem 2.2 If A and B are disjoint convex sets

in a linear space £, then there exist complementary

convex sets C and D of £ such that A^C and B^D.

Proof: Consider the class P={(A^,B^)}, where

A^ and B^ are convex sets in £ such that A flBj_=(j), A^A^, (

B^B^. P is not empty since (A,B) e P. Now define a

partial order on P as follows:

(Ai,Bi)<(Aj,Bj) if A ^ A j and B^Bj.

The union of every linearly ordered subset of elements

in P belongs to P. By Zorn's Maximal Principle there

exists a maximal element (C,D) in P.

Now all that remains is to show that C=D. :

Suppose that p e £^(CuD) . By Theorem 1.7

conv(Cu{p})= u [aC+(1-a){p}] and also 0 < a < l

19

conv(Du{p})=0<u<;L[aD+(l-a) {p}] . Since (C,D) is maximal,

there exist points x e Dftconv(Cu{p}), y e Cftconv(Du{p}),

such that x fj: C and y £ D. But the above implies that

there exist points c e C, d e D such that x e intvcp,

y e intvdp. The points c, d, p determine a triangle'and,

like medians, dxQcy tf). Since dx^D, cy^C, then C£JD (f>,

a contradiction. Thus C=D. »

Theorem 2.3 If f and g are linear functionals in

I such that [f :a] = [g:3] , then there exists a constant X^O

such that f=Xg, a=X3-

Proof: If f or g is identically zero, the theorem

is trivial. Suppose f^O and g|0. If 3=0, choose

z e &Mg:3]. If 3^0/ choose z e [g: 3 3 - Thus g(z)?*0.

Let p e Si. Then there exists a t e [g:3] such that p=t+yz,

where y e R. Now, f(p-t)=yf(z), and g(p-t)=yg(z). Then

g(p-t)/g(z)=y so that f(p-t)=[f(z)/g(z)][g(p-t)]. Let

X=f(z)/g(z). Thus f (p) -f (t) = Xg(p)-Xg (t) , and hence

f(p)-a=Xg(p)-X3. Since the previous statement is true

for all p e l , then it is true for.p=z. Consequently,

a-X3=f(z)-Xg(z)=0. Hence a=X3« Immediately, f(p)=Xg(p).

Definition 2.6 A hyperplane H is said to support

a set S at a point x e S iff x e H and H bounds S.

Theorem 2.4 Suppose A and B are convex subsets of

a linear space £, and that coreB fj), A <J>, and AftcoreB=4>.

Then there exists a hyperplane H=[f:a] such that f(A)<a

and f(B)>a.

20

Note that in a finite-dimensional linear

topological space "core" corresponds, to "interior."

For a proof of Theorem 2.4, which is a fundamental

separation theorem, see Valentine (3, p. 24).

Theorem 2.5 Suppose that B is a convex body

in a topological linear space L, and suppose that F

is a flat in L such that FfiintB=<J>. Then there exists

a hyperplane H containing F which bounds B.

Proof: By Theorem 2.4 there exists a hyperplane

H*=[f:a] such that f(F)<ot and f(B)>a. Then there

exists a translate H=H*+x of H* such that F?H and

H bounds B.

Theorem 2.6 A hyperplane H=[f:a] in L bounds a

nonempty open set iff f is continuous and'f|0.

For a proof of Theorem 2.6 consult Valentine

( 3, pp.25-26).

Theorem 2.7 A hyperplane H=[f:a] is closed iff

f is continuous with f^O.

Proof: Suppose H is closed. Since H^L, there

exists a point x £ H and since H is closed, x $ clH so

that there exists a neighborhood N(x) of x such that

N(x)QH=<f>. Now by Theorem 2.6, f is continuous since H

bounds N(x).

Suppose that f is continuous with f£0. Then

{ x | f(x)<a } and { x | f(x)>a } are open convex sets and

H=L^[{x|f(x)<a}u{x|f(x)>a}] so that H=[f:a] is closed.

21

Theorem 2.8 Suppose that C is a closed convex

set in a locally convex space L and that x e I>C.

Then there exists a closed hyperplane H through x such

that HflC=4 .

Proof: IAC is a neighborhood of x. Then there

exists a convex neighborhood N(x) of x such that

N(x)CL^C and N(x)ftC=<}> since L is Hausdorff, locally

convex, and C is closed. The interior of N(x) <}>, so

that Theorem 2.4 implies the existence of a closed

hyperplane H* separating N(x) and C. Now consider the

translate H of H* such that x e H. HflC=4>.

Theorem 2.9 If C is a convex body in a topological

linear space L, then through each bour*dary point of C ,,

there passes a closed hyperplane of support.

Proof: If C=L the conclusion is trivial. Suppose

that C^L. Let x e bdC, and note that {x} is a convex

set. By Corollary 1.11 intC is convex. But Theorem 2.4

implies that there exists a hyperplane H* which separates

intC and {x}. Now consider a translate H of H* such that

x e H. Then H is a hyperplane of support at x.

Theorem 2.10 Suppose that S is a set in a

topological linear space L and that int convS <{>. A point

p e int convS iff each hyperplane H through p strictly

separates at least two points of S.

Proof: Suppose that p e int convS and that there ?

exists a hyperplane H=[f:a] such that p e H and H does

22

not separate at least two points of S. But this

statement means that all points of S belong to a closed

half-space determined by H, say {x|f(x)<.a}, since f is

continuous by Theorem 2.6. But now it is apparent that

p e bd convS, a contradiction. Thus each hyperplane

through p separates points of S.

Now suppose that each hyperplane through p

strictly separates at least two points of S. And further

suppose, by way of contradiction, that p ^ int convS.

By Theorem 2.5 and Theorem 2.7, since int convS <J>, there

exists a closed hyperplane through p bounding S, a

contradiction. Thus p e int convS.

Theorem 2.11 A line through the interior of a

compact convex body A intersects bdA twice.

Proof: Let p e intA tj). Since A is compact, A^L.

Consider a ray Q from p. A is bounded, but Q is

unbounded so that there exists some point y e (L'vA)QQ.

Now let Q={x e L | x=ay+(l-a)p, a>0 }. Consider line

segment py={ x e Q | O^a^l }. Order the elements of py

as follows: if w e py, z e py, where w=3y+(l-g)p and

z=Ay+ (1-A) p for some CK3, A<1, then z>w iff A>$. The

pyQA is bounded above by y. Then there exists a least

upper bound k for pyflA. Obviously, k e bdA and also

k e QfibdA. Similarly, for a ray Q*={x e L | x=ay+(l-a)p,

a<0} the same conclusion as before is obtained. So that

QuQ* is a line through p and QoQ* intersects bdA in at

23

least two points. In addition, to intersect in more

than two points contradicts the convexity of A.

Theorem 2.12 Let S be a closed set in a

topological linear space L, and suppose that intS <}>.

Then S is convex iff through each boundary point of S

there passes a hyperplane of support to S.

Proof: The necessity is provided immediately

by Theorem 2.9.

Now suppose that through each boundary point of

S there passes a hyperplane of support to S. If S=L,

then S is convex. Suppose S^L. Then let x e intS and

y e I/vStj). As in Theorem 2.11, there exists a z e bdS

such that z e xy. By hypothesis, there exists a

hyperplane of support H=[f:a] through z. y £ H, for if

y e H, then x e H since x, y, and z are colinear. But

x e H would contradict that H is a hyperplane of support.

Thus the closed half-space H+={x|f(x)>a} determined by

H and containing x contains S but not y where y is any

point in the complement of S. Let T be the class of

all such half-spaces that contain S but no point y

external to A. Thus, OT=S. By Theorem 1.4, S is convex

since each element of T is convex.

A finite dimensional normed linear space is sometimes

called a Minkowski space. Consult the Appendix for a more

formal definition of a Minkowski space. For a proof of

the following theorem see Valentine (3, p. 40).

24

Theorem 2.13 If S is a compact set in a Minkowski

n

space L , then convS is compact.

The closed convex hull of a compact set is also

compact in a Ban'ach space. A Banach space is a normed

linear space in which the Cauchy criterion is sufficient

for the convergence of sequences. For a proof of

Theorem 2.13 in Banach spaces see Taylor.(2).

•CHAPTER BIBLIOGRAPHY

1. Kelley, J. L. , General Topology, Princeton, N. J., D. Van Nostrand Company, Inc., 1955.

2. Taylor, A. E., Introduction to Functional Analysis, New York, John Wiley & Sons, Inc., 1958.

3. Valentine, Frederick A., Convex Sets, New York, McGraw-Hill Book Company, 1964.

25

CHAPTER III

EXTREMAL CHARACTERIZATION OF CONVEX SETS

Sufficient notions have now been introduced to

enable an examination of some sets which share a convex

hull, in particular, those sets which are minimal and

generate a given convex hull. The theorems in this

section culminate in extremal characterization of

convex sets including the Krein-Milman Theorem and

other related results.

Definition 3.1 A point x e bdC, where C is

convex arid Cc;L, is called an exposed point of C iff there

exists a hyperplane of support H to C through x such

that HflC={x}.

Rather than introduce a norm in order to define a

n

Minkowski space L , the following theorem is offered

without proof and only the comment that En is a Minkowski

space so that the theorem is intuitively correct. For

a proof consult Valentine (2, p. 52).

Theorem 3.1 Let C be a closed convex set in a

Minkowski space L2 (E2). Each compact connected portion

of the boundary of C which is not contained in a line

26

' 27

segment contains an exposed point of C. If C is a line

segment, it has two exposed points.

Definition 3.2 If S is a convex set in Z, then • *

a point x e S is an extreme point of S iff no non- ' 1

degenerate segment in S exists which contains x in its

relative interior.

Note that x is an extreme point of a convex set S

iff S^{x} is convex. Also, the above definitions imply

that the exposed points of S are extreme points of S.

Definition 3.3 A non-empty set M of a set K£L

is called an extremal subset of K iff x e K, y e K,

and Mf2intvxy <}> implies that {x}u{y}£M.

An extreme point of a set K is an extremal subset

of K consisting of just a single point.

Definition 3.4 If C is a compact convex set of

a topological linear space L, then extC denotes the set

of extreme points of C and expC denotes the set of

exposed points of C.

Definition 3.5 A real valued function is often

called a functional. Let p be a function defined on a

subset of Z. Then p is a convex functional defined on

a convex subset S of £, iff

p[ax+(1-a)y]=ap(x)+(1-a)p(y)

holds for all x, y e S and 0<a<l.

28

Definition 3.6* Let A bo a compact convex subset

of LU. Then the set denoted by gA={x | x£A and convx=A)

is called the genesis of A.

The elements of gA are called the generators of A.

The set y e gA such that y£x for every x e gA ( if such

a set exists ) is called the antecedent of A. If gA is •

considered to be partially ordered in the usual sense • <

of partially ordering sets, then the antecedent of A is

a minimal set analogous to Zorn's maximal element.

The following examples provide some clue as to

the nature of sets having antecedents:

Example 3.1 The antecedent of a closed line

segment is the set of its endpoints.

Example 3.2 The antecedent of a closed triangular

region is the set of its vertices. Further, the

antecedent of a solid polyhedron is the set of vertices

of the polyhedron.

Example 3.3 The antecedent of A={x c En: Ix-pl^r}

is the set {x e ,En: |x-p|=r}.

Example 3.4 Consider the open ball A where

A={x e En: |x-p|<r}. Although A is not closed and hence

not compact, it is informative to note that there are

infinitely many subsets of A which share A as a convex

hull, but no smallest such set. On the other hand, the

set B={(x,y) e E2| y>x2} is unbounded and thus not

29

compact but has a smallest set {(x,y) e E2| y=x 2H B

whose convex hull is B.

Notice that in examples 3.1 - 3.3 the antecedent

of each convex set was the set of exposed points of the

convex set. One might be led to conjecture that the

exposed points of a compact convex set in Ln is the

antecedent of the compact convex set. Example 3.5

quickly dispels this notion.

Example 3.5 Let A={x e E2: |x-s|=r} and p e E2^A.

Let B=conv(Au{p}). The antecedent of B is not the set

of exposed points of B. Let pw and pz be the tangents

from p to circle A where w, z e A. The antecedent of B

is the set {closed major arc zw}u{p}. But z and w are

not exposed points of B. They are, however, extreme

points of B.

From the definitions of expC and extC, it is

obvious that expCi;extC. But Example 3.5 clearly shows

that extC is not a subset of expC. However, both extC

and expC are subsets of bdC.

Theorem 3.2 If the antecedent of a compact

convex set A Ln exists, it is unique.

Proof: Suppose x is the antecedent of a compact

convex set A. Now suppose, by way of contradiction,

that there exists a y e gA such that y is an antecedent

for A and y^x* But by definition of antecedent x?y and

30

yCx. Then y=x, a contradiction. So that if the

antecedent of A exists, it is unique.

Theorem 3.3 If F is a family of extremal

subsets of a set K, then a non-empty intersection of

any subfamily of F is an extremal subset of K.

Proof: Let f^ e F, i e A. Choose x e K, y e K.

If ( Q f.) fJintvxy cj), then f. f2intvxy?*(j>, so that for ieA i i

each f•, {x}u{y}?f•, i e A. Hence, {x}u{y}£ Q f.. i i i^A 1

Definition 3.7 * Let S be a set. Let T be a

collection of subsets of S, then (S,T) is said to be

a restricted topological space iff for every non-empty

subcollection F<;T

(a) uF e T and

(b) if F is finite, ftF E T.

Theorem 3.4 The genesis of a compact convex set

having an antecedent is a restricted topological space.

Proof: Let F be a finite subset of gA where A

is a compact convex subset of Ln such that A has an

antecedent. Let x be the antecedent of A. Obviously,

xcy for every y e F. Then x£ftF£A. But, convx^conv(QF)

and conv(AF)£A. Now convx=A, so that conv(QF)=A. But

this implies that fiF e gA. A similar argument produces

that if F*CgA, then uF* e gA.

31

Theorem 3.5 If K is a non-empty compact set in

a locally convex topological linear space L, then K has

at least one extreme point.

Proof: Let F be the collection of all compact

extremal subsets of K. Then F is not empty since KCF.

Partially order the collection F by set inclusion.

Since K is compact, any non-empty linearly ordered

subset of F has a compact non-empty intersection, which

by Theorem 3.3( is an extremal subset of K. By Zorn's

Maximal Principle, the set F has a minimal element,

denoted by F*.

Now all that remains is to show that F* is an

extremal subset of K consisting of exactly one point

and thus an extreme point of K. Suppose, by way of

contradiction, that F* contains two distinct points

x and y. Now {x} is a closed convex set and y e IA>{X}.

By Theorem 2.8, there exists a closed hyperplane H such

that y e H and HJHx}=<f>. Then x £ H and F*ftHj*<j>. Since

F*QH is clearly an extremal subset .of K, and since

x £ F*ftH, the set F* is not minimal. Hence F* can

contain at most one point. This statement implies that

F*e K is an extreme point of K, since an extreme point

is an extremal set consisting of exactly one point.

Theorem 3.6 Let S be a compact convex set in a

locally convex topological linear space L. Then for each

closed hyperplane of support to S, HfiextSj <j>.

32

Proof: Let H be a closed hyperplane of support

to S. HQS <j>, where H=[f:a] and S is a subset of one of

the half-spaces determined by H, say H+. Since H is

closed, HQS is a non-empty, compact convex set and by

Theorem 3.5 HQS has an extreme point x. If y e S'v-H,

z e S, then f(y)<a, f(z)jca, where without loss of

generality H+={x e L| f(x)<a }. Since*f(x)=a, then

x £ intvyz. Also, if y e HQS, z e HQS, then since

x is an extreme point of SQH, x £ intvyz. Thus x is

an extreme point of S.

The Krein-Milman Theorem, which follows, is a

well known theorem that describes a convex set in terms

of its extreme points. For a different proof of

Theorem 3.7 see Krein and Milman (1).

Theorem 3.7 If S is a compact set in a locally

convex topological linear space L, then the closed

convex hull of S is the same as the closed convex hull

of the set of extreme points of S.

Proof: S Lnce extS£S, then cl conv extS is a

subset of the closed convex hull of S. Now it remains

to show that S£cl conv extS. Suppose, by way of

contradiction, that S is not a subset of cl conv extS,

and then choose x from S^cl conv extS cJ). Since L is

a locally convex space, Theorem 2.8 implies that there

exists a closed hyperplane H*=[f:B] through x, where

33

without loss of generality, f(y)<$ for y e cl conv extS.

But S is compact, so that a hyperplane H=[f:a], a>3,

exists such that HftS <|>, f(S)<a, and thus HQcl conv extS=<f>.

But Theorem 3.6 'implies that since H is closed, Ii

contains an extreme point of S. This is a contradiction.

Thus, it is false that S is not a subset of cl conv extS.

Then cl conv S= cl conv extS.

Theorem 3.7 may be restated as follows:

Theorem 3.8 Suppose that C is a compact convex

subset of a locally convex Hausdorff linear space L and

x is a subset of C. Then C is the closed convex hull

of x iff each extreme point of C lies in the closure of x.

A finite dimensional normed linear space Ln is

locally convex, so that Theorems 2.13, 3.7, and 3.8 now

imply that every compact convex subset of Ln has an

antecedent, namely the set of extreme points of the

compact convex set. Recall also that Ln is linearly

isomorphic to En. These statements then form the proof

of the following:

Theorem 3.9 If S is a compact convex set in a

finite dimensio-nal normed linear (Minkowski) space Ln

or En, then S has an antecedent.

CHAPTER BIBLIOGRAPHY

1. • Krein, M. G. and D. P. Milman, "On Extreme Points of Regular Convex Sets," Studia Mathematica, IX (1940), 133-138.

2. Valentine, Frederick A., Convex Sets, New York, McGraw-Hill Book Company, 1964.

34

CHAPTER IV

A PROPOSITIONAL SUBSET OF

THE GENESIS OF A SET

An advantage of defining the genesis of a compact

convex subset of Ln is to discuss some properties of

some of the propositional subsets of the genesis. As a

case in point, the following definitions l^ad to one

such subset of gA, where A£E3:

Definition 4.1* If X is a set property and A

is a compact convex subset of Ln, then PA(X) will be

used to denote the subset of gA consisting of those

elements having property X, i.e., PA(X)={y e gA|y has X}.

Definition 4.2* Let A be a subset of L, then the

set denoted by augA= u xy is the augmentation of A. xeA yeA

Definition 4.3 A maximal connected subset of a

set A in a topological space S is called a component of A.

A set A in L is simply connected iff each component

of the complement of A is unbounded.

Alternately, a set is simply connected iff any

closed curve within it can be deformed continuously to

a point of the set without leaving the set.

35

36

Definition 4.4* Let A£Ln, A not convex. Suppose

x e convA^A and that there exists points z and y

belonging to A such that x=az+(l-a)y for some 0<a<l,

then x is said to be braced relative to A. zy is an

A-relative brace of x. If every point of convA^A has

an A-relative brace, then A is said to be a braced set.

Definition 4.5 A crosscut of a set S^L is a

closed segment xy such that intvxy^intS and such that

x e bdS and y e bdS.

Theorem 4.1 If an open set K has no crosscuts,

then it i-s the complement of a convex set.

Proof: Let x, y £ K. If xy£2K <j>, then K has a

crosscut which is a subinterval of xy since K is open.

Hence xy K=(f> and K is convex.

Definition 4.6 If S is a set in a linear space I

and if V is a K-dimensional flat, then SflV is called a

K-dimensional section of S. If H is a hyperplane of I,

then SQH is called a plane section of S.

Definition 4.7 A continuum in a topological

linear space L is a compact connected set. A closed

connected set in a Minkowski space Ln is called a

generalized continuum . A generalized continuum is

automatically boundedly compact.

Definition 4.8* If A£L3 and A is not convex and

x e convA^A implies that there exists a plane H such

37

that x c H and AilH is connected and x e conv(AflH),

then A is said to be rigid.

Note that the set of braced sets whose convex

hull is convA is a propositional subset of g(convA).

If B is the property of being braced and C is a convex

set then The following lemma is presented

preparatory to determining a sufficient condition for

a set in L2 to be braced. Because of Theorem 1.3, a

proof in E2 is adequate.

Lemma 4.1 ; If A is a connected subset of E2

and p e A and q e augA^A and x=ap+(l-a)q for 0<a<l,

then x e augA.

Proof: If x e A, the result follows immediately.

Suppose x £ A. q e augA^A iff there exist 0<3<1 and

w, z e A such that q=f3w+(l-3)z. Let

[f:a]={y e E2| y=Ap+(l-A)q, A e R},

[g:S]={y e E2| y=Az+(l-A)x, A e R},

[h:y] = {y e E2 | y=Aw+(l-A)x, A e R} .

Note that if f(w)>a, then f(z)<a.

If g(p)>6, then g(q)<6, g(w)<6.

If h(p)>y, then h(z)<y, h(q)<y.

The above statements may be adopted without loss

of generality.

Clearly, p and w are elements of a connected set

A separated by Cgx 63 . Then D=AQ [g: otherwise A is

38

a subset of the union of the two open half-planes

determined by [g:6], contradicting that A is connected.

Case I: Suppose T^=Dft{x| f (x) Let t^ e T^.

Then there exists 0<A<1 such that x=Az+(1—A)t^. But

t^ e A, thus x e augA.

Case II: Suppose that T =4>. Then Dfi[f:a] (J) or

T2=Dfi{x | f (x) <a}j(j). But DQ[f:a] <{> implies that x e A,

a contradiction. Thus T2 <J>. Let t2 e T2" Then there

exists 0<A<1 such that t2=Az+(1-A)x. Thus h(t2)<Yr and

t2 and p are separated by [h:y].

Let £ T =Aft [h: y] ft{x | f (x) <a}fi{x | g (x) >6 . Then

t =Ax+(l-A)w where A>1. Now x=(1/A)t--t(1-A)/A]w, and 3

x=-[(l/A)-l]w+(l/A)t3. Let A=l/A. Then 0<A<1 so that

x=Xt^+(l-X)w. The last statement implies that

x e augA.

Theorem 4.2 If A is a compact connected subset

of E 2, then augA=convA.

Proof: Obviously augA<;convA.

Now suppose that x e convA. Since A is compact,

if x e bd convA, then x e extA^A or x belongs to the

relative interior of some maximal line segment zy where

zy^bd convA. Now line zy bounds A so that by Theorem 3.6

and Definition 3.2, z, y e extA^A. Thus if x e bd convA,

then x e augA.

39

Suppose that x e int convA. Let p be an extreme

point of A, thus p e bd convA. Consider the line

q=Ax+(1-A)p. qftint convA cj> so that by Theorem 2.11,

qflbd convA^{p} consists of exactly one point, say k.

Note that q=Ak+(l-A)p and for some 0<A<1, x=Ak+(l-A)p.

If k e A, then x e augA.

If k £ A, k e augA and by Lemma 4.1, x e augA.

Thus, in either case, convA^augA, so that augA

and convA are identical, i.e., augA=convA.

It is now obvious that if A is a compact connected

subset of L2, then augA is the convex hull of A.

Theorem 4.3 now follows immediately.

Theorem 4.3 If A is a compact connected subset

of L2 or E2, then A is braced.

Observe also that Theorem 4.2 together with

Theorem 1.7 implies the following:

Theorem 4.4 If A and B are disjoint subsets of

E2, and A is a compact connected set and B is a compact

connected set, then AUB is braced.

One should not be too hasty and conclude that

connectedness is sufficient for a compact set to be

braced in L3. The following example illustrates:

Example 4.1 Let A be a subset of E3 formed by

taking the union of three non-coplanar line segments

having a common point of origin. A is compact and

40

connected, but no point in the interior of the closed

solid tetrahedron which is A's convex hull has a

brace relative to A.

Other examples may be easily constructed to show

that connectedness is not necessary for.a set to be

braced in E2 or E3 and consequently L2 and L3.

For a proof of the following theorem consult

Valentine (1, pp. 91-93).

Theorem 4.5 If S is a compact set in L3 and if

each plane section of S is a simply connected continuum,

then S is convex. (The converse is obviously true.)

Theorem 4.6 If A is a subset of E3 or L3 and

A is rigid, then A is braced.

Proof: The theorem follows immediately from the

definition of rigid and Theorem 4.3.

The results of Chapter IV are now summarized in

terms of. Definition 4.1.

Theorem 4.7 Let B be the property of being braced.

Let C be the property of being connected. Let R be the

property of being rigid. Now suppose that X is a convex

subset of L2 and Y is a convex subset of L3, then

PX(CKPX(B) and Py(R)cPy(B).

Proof: The results are immediate from Theorem 4.2

and Theorem 4.5.

CHAPTER BIBLIOGRAPHY

1. Valentine, Frederick A., Convex Sets, New York, McGraw-Hill Book Company, 1964.

41

APPENDIX

Definition A.l A set S^i, star-shaped with

respect to the origin Q, is called linearly bounded iff

each line through Q intersects S in a line segment.

Definition A.2 Let S be an open set in L which

is star-shaped with respect to 0 and which is linearly

bounded. Then the Minkowski distance functional p is

a real valued function defined on L as follows:

p(x)=A>0 where \x*=x, x* e bdS, and ax*^intS, 0<a<l.

If x e S, then p(x)£l.

Definition A.3 A topological linear space L is

said to be normable if it is locally convex and if it

contains a non-empty bounded open set.

Definition A.4 A normable topological linear

space L with a norm (as for example, the Minkowski

distance functional) is called a normed linear space.

Definition A.5 A finite-dimensional normed '!

linear space is called a Minkowski space.

AO

BIBLIOGRAPHY

Books

Kelley, J. L., General Topology, Princeton, N. J., D. Van Nostrand Company, Inc., 1955.

Taylor, A. E., Introduction to Functional Analysis, New York, John Wiley & Sons, Inc., 1958.

Valentine, Frederick A., Convex Sets, New York, McGraw-Hill Book Company, 1964.

Articles

Krein, M. G. and D. P. Milman, "On Extreme Points of Regular Convex Sets," Studia Mathematica, IX (1940), 133-138.